# Expansion of an integral

I have an integral of the form $$I=\int_0^{\infty}{dx}\ln \bigg(1+\exp(-\frac{f(x)}{a})\bigg)$$ where $$a$$ is a positive constant and $$f(x)$$ is a regular and positive function such that $$I$$ is finite (for example: $$f(x)=x$$); moreover $$f(x)=x+O(x^3)$$ for $$x \rightarrow 0$$. I have to expand this integral $$I=c_0+c_1a+..+c_na^n+O(a^{n+1})$$ for $$a \rightarrow 0^+$$, with the value of the integral beeing dominated by the behaviour of $$f(x)$$ near $$0$$. Can anyone give me some hints or references to compute the first coefficients of this expansion?

• Does the integral converge at infinity? Does not look so. – Fedor Petrov Feb 21 at 17:24
• I've edited the question, $f(x)$ is such that $I$ is finite – John Feb 21 at 17:30
• how can it be finite? – Fedor Petrov Feb 21 at 18:26
• It is an hypothesis. – John Feb 21 at 18:29
• I've edited my question: f(x) is positive and $I$ finite. An example: $f(x)=x$. The point is: how can i do the expansion? I had in mind this example: f(x)=x*g(x), where g(x) is posivite and bounded – John Feb 21 at 18:36

Let $$f(x)$$ have the expansion $$f(x)=x+c_3 x^3+c_4 x^4 +\cdots$$, then define $$y=x/a$$ and you have $$I=\int_0^{\infty}{dx}\ln \bigg(1+\exp\left(-\frac{f(x)}{a}\right)\bigg)$$ $$=\int_0^\infty \left[a\ln \left(1+e^{-y}\right)-\frac{a^3 c_3 y^3}{e^y+1}-\frac{a^4 c_4 y^4}{e^y+1}+{\cal O}(a^5)\right]\,dy$$ $$=\frac{1}{12} \pi ^2 a-\frac{7}{120} \pi ^4 c_3 a^3-\frac{45}{2} \zeta (5)c_4 a^4+{\cal O}(a^5).$$ The term of order $$a^p$$ has coefficient $$-c_p\left(1-2^{-p}\right)p! \zeta (p+1)$$, when $$c_p$$ is the coefficient of order $$x^p$$ in the expansion of $$f(x)$$.